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Theorem: Consider the metric space $(X, d)$, where $X \subseteq \mathbb{R}$ and $d(x, y) = |f(x) - f(y)|$, with $f$ being a one-to-one (injective) function on its domain $X$. If the range of $f$ is a closed set in $\mathbb{R}$, then the metric $d$ induces a complete metric space.

Proof: Let $x_n$ be a Cauchy sequence in the metric space $(X, d)$. By the definition of the metric $d$, we have $d(x_n, x_m) = |f(x_n) - f(x_m)| \to 0$ as $m, n \to \infty$ since $f$ is a one-to-one function.

Now, consider the real numbers with the usual metric. The sequence $f(x_n)$ is a Cauchy sequence in $\mathbb{R}$ (using the standard metric), and since $\mathbb{R}$ is complete, $f(x_n)$ converges to a limit $l = f(a)$ for some $a \in X$(using closeness of range).

Next, we show that $x_n$ converges to $a$ in the metric space $(X, d)$. Consider $d(x_n, a) = |f(x_n) - f(a)|$, and as $f(x_n) \to f(a)$, we have $d(x_n, a) \to 0$ as $n \to \infty$. Therefore, $x_n$ is a convergent sequence in the metric space $(X, d)$.

Since every Cauchy sequence in $(X, d)$ converges to a point in $X$, we conclude that the metric space $(X, d)$ is complete.

My questions are as follows:

$1.$ Is the statement and proof of the theorem correct?

$2.$ If everything is correct, can we make further improvements, such as turning the theorem into an "if and only if" statement?

Please suggest. Thank you .

neelkanth
  • 6,550

2 Answers2

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The statement is correct. In the proof, $d(x_n,x_m)=|f(x_n)-f(x_m)| \to 0$ as $m,n \to \infty$ by definition of the metric $d$, not because $f$ is injective. Injectivity of $f$ is only needed to ensure that $d$ is a metric, i.e. $d(x,y) = 0$ implies $x = y$. Otherwise $d$ would only be a pseudometric but the statement would also be correct.

The converse is also true: if $(X,d)$ is complete then $f(X)\subseteq \mathbb{R}$ is closed. For, let $x_n \in X$ and $f(x_n) \to y$ in $\mathbb{R}$. Then $(f(x_n))$ is a Cauchy sequence in $\mathbb{R}$, so $d(x_n,x_m) = |f(x_n)- f(x_m)| \to 0$ as $n,m \to \infty$. Since $(X,d)$ is complete, there is some $a \in X$ with $x_n \to a$ in $(X,d)$. So, $|f(x_n)-f(a)| = d(x_n,a) \to 0$ as $n \to \infty$. Now $f(x_n) \to y$ and $f(x_n) \to f(a)$ as $n \to \infty$, so $y = f(a)$ and therefore $y \in f(X)$. Thus $f(X)\subseteq \mathbb{R}$ is closed.

psl2Z
  • 5,772
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psl2Z already answered this fine, let me just augment with another way to understand what’s going on.

By definition, $f\colon (X,d)\to (f(X),d_e)$ is an isometry, where $d_e$ is the Euclidean distance. Therefore, $(X,d)$ is complete if and only if $f(X)$ is complete in the Euclidean metric. But $\mathbb R$ is complete, hence $f(X)$, as a subset of $\mathbb R$, is complete if and only if it is closed.

M W
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